When a multiplication operator $M_\phi$ satisfies in ${M_\phi}^{2}=M_\phi$?

Question

Let $(X,\Omega,\mu)$ be a $\sigma$-finite measure space and put $\mathscr{H}=L^{2}(X,\Omega,\mu)=L^2(\mu)$. If $\phi \in L^\infty(\mu)$, define $M_\phi:L^2(\mu) \to L^2(\mu)$ by $M_\phi f=\phi f$. Then $M_\phi \in \mathscr{B}(L^2(\mu))$ and $\|M_\phi\|=\|\phi\|_\infty$.

The operator $M_\phi$ is called a multiplication operator. I want to show that

$M_\phi=M_\phi^2$ if and only if $\phi$ is a characteristic function.

$\longleftarrow$ If $\phi$ is a characteristic function we have $M_\phi(M_\phi(f))=M_\phi(\phi f)=M_\phi(f)$.

How I can prove the converse?

Answer 1

We want to show that if $\phi \in L^\infty$ and $\phi f = \phi^2 f$ for every $f$ in $L^2$ then $\phi$ must be a characteristic function. That is, we want to show that $\phi$ takes values in $\{0,1\}$ a.e. (since $\phi$ is measurable by assumption).

Suppose that this is not the case. Then one of the measurable sets $A_1 = \{ x: \phi(x) > 1 \}, A_2 = \{ x: \phi(x) \in (0,1) \} , A_3 = \{ x: \phi(x) < 0\}$ has positive measure. Say $\mu(A_i) > 0$. Then since our measure space is $\sigma$-finite, $A_i$ has a measurable subset of positive, finite measure which we call $B_i$. Then $1_{B_i} \in L^2$. In particular, $\phi 1_{B_i} = \phi^2 1_{B_i}$ almost everywhere. But this is impossible by construction of the $B_i$.

Answer 2

$M_\phi^2= M_\phi$ holds iff $\phi^2 f = \phi f$ for all $f \in L^2$. Since $\mu$ is $\sigma$-finite there is a sequence of subsets of finite measure $E_n$ such that $X = \bigcup_n E_n$, so in particular $\phi^2 \chi_{E_n} = \phi \chi_{E_n}$ i.e. $\phi^2 = \phi$ on $E_n$ for every $n$. Hence $\phi^2 = \phi$ or $\phi(\phi - 1) = 0$ so $\phi$ must be $\{0,1\}$-valued.

Answer 3

This is really the same as what nobody said, rephrased so as to point out the following basic fact:

Lemma If $\mu$ is $\sigma$-finite then $||M_\phi||=||\phi||_\infty$.

It's clear that $||M_\phi||\le||\phi||_\infty$ for any measure. Suppose $\mu$ is $\sigma$-finite and $\alpha<||\phi||_\infty$. Let $S=\{x:|\phi(x)|\ge\alpha\}$. By definition $\mu(S)>0$. So there exists $E\subset S$ with $0<\mu(E)<\infty$. If $f=\chi_E$ then $||\phi f||_2\ge\alpha||f||_2>0$, hence $||M_\phi||\ge\alpha$.

So $||M_\phi||\ge\alpha$ for every $\alpha<||\phi||_\infty$, hence $||M_\phi||\ge||\phi||_\infty$.

And now you're done: $M_\phi^2=M_\phi$ if and only if $M_{\phi^2-\phi}=0$; the lemma says this happens if and only if $||\phi^2-\phi||_\infty=0$, which says $\phi^2=\phi$ almost everywhere.